tag:blogger.com,1999:blog-11295132.post112259701861393027..comments2018-02-12T01:50:07.787-08:00Comments on A Neighborhood of Infinity: Automatic DifferentiationDan Piponihttps://plus.google.com/107913314994758123748noreply@blogger.comBlogger10125tag:blogger.com,1999:blog-11295132.post-67808706756496689032013-06-27T18:13:57.457-07:002013-06-27T18:13:57.457-07:00Anyone looking for the paper can find it on the Wa...Anyone looking for the paper can find it on the <a href="http://web.archive.org/web/20120111180418/http://homepage.mac.com/sigfpe/paper.pdf" rel="nofollow">Wayback Machine</a>.Jeffnoreply@blogger.comtag:blogger.com,1999:blog-11295132.post-70636335999597051772013-06-12T07:32:21.977-07:002013-06-12T07:32:21.977-07:00The link to your paper is brokenThe link to your paper is brokenAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-11295132.post-48925127518441410272013-06-12T07:32:04.633-07:002013-06-12T07:32:04.633-07:00The link to your paper is brokenThe link to your paper is brokenAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-11295132.post-61346346693964538912013-06-12T07:31:27.237-07:002013-06-12T07:31:27.237-07:00Link to your paper is broken.
Very nice blog and ...Link to your paper is broken.<br /><br />Very nice blog and repliesAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-11295132.post-51247761136987462602007-07-26T03:37:00.000-07:002007-07-26T03:37:00.000-07:00I thought so, or we wouldn't have the 800 differen...I thought so, or we wouldn't have the 800 different methods for numerical integration :P<BR/><BR/>What do you get if you *do* integrate with dual numbers? I guess I'll go have a peek at Wikipedia.Porgeshttps://www.blogger.com/profile/02727258157936734796noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-87470349354676470042007-07-25T18:20:00.000-07:002007-07-25T18:20:00.000-07:00porges,There isn't a corresponding method for inte...porges,<BR/><BR/>There isn't a corresponding method for integration. Integration turns out to be very ugly compared to differentiation.sigfpehttps://www.blogger.com/profile/08096190433222340957noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-5781098171934200082007-07-24T19:44:00.000-07:002007-07-24T19:44:00.000-07:00Neat. Is there something similar for integration?Neat. Is there something similar for integration?Porgeshttps://www.blogger.com/profile/02727258157936734796noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-87888542408628066212007-02-27T04:58:00.000-08:002007-02-27T04:58:00.000-08:00I was also confused by the f(x + d ) = f(x) + f'(x...I was also confused by the <BR/>f(x + d ) = f(x) + f'(x)d thing.<BR/><BR/>I played with it a while, and this more verbose derivation may help those of us with less experience.<BR/><BR/>f(x) = c0 + c1*x + c2*x^2 + ... + cn*x^n<BR/><BR/>therfore:<BR/>f(x + x'*d) = c0 + c1*(x+x'*d) + c2*(x+x'*d)^2 + ... + cn*(x+x'*d)^n<BR/><BR/>= c0 + c1*x + c2*x^2 + ... + cn*x^n<BR/>+ c1*x'*d + 2*c2*x*x'*d + ... + n*cn*x^(n-1)*x'*d<BR/><BR/>= f(x) + f'(x)*x'*d<BR/><BR/>By the way, this is one of the best blogs I've read in quite a while. I'm basically done with my undergrad math degree, and this is interesting stuff. It's real applications of all the theory i picked up in abstract algebra.<BR/><BR/>Thanksshanealhttps://www.blogger.com/profile/10780697234154191381noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-1154318343725689342006-07-30T20:59:00.000-07:002006-07-30T20:59:00.000-07:00Hi Suresh,That identity doesn't hold for real numb...Hi Suresh,<BR/><BR/>That identity doesn't hold for real numbers but in an extended algebraic structure where dÂ²=0. If you look at my own <A HREF="http://homepage.mac.com/sigfpe/paper.pdf" REL="nofollow">paper</A> on the subject you'll see some actual C++ code that implements numbers in such a structure. That should bring it down to earth in a practical way.sigfpehttps://www.blogger.com/profile/08096190433222340957noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-1154312501768376002006-07-30T19:21:00.000-07:002006-07-30T19:21:00.000-07:00i guess I don't get it. how isf(x + d ) = f(x) + f...i guess I don't get it. how is<BR/><BR/>f(x + d ) = f(x) + f'(x)d ?Sureshhttps://www.blogger.com/profile/16443460499476270978noreply@blogger.com